Divide contents of a unsigned char array into 2 halves - c

Question

I am wondering if anyone can help me I am only learning c, I am trying to Divide contents of a unsigned char array into 2 halves, which the result can stored in two unsigned int's,
For example purposes, I have some code below which adds a hex value into a BYTE array, so How would split the contents of val[] into two but keep the same order
#include <stdio.h>
typedef unsigned char BYTE;
int main()
{
// Sample purposes putting hex into val[8]
int i,j;
long long hex=0x78661EB54FE76763;
BYTE val[8];
for(j=0,i=7; i>=0; i--,j++){
val[j]= (hex>>(i*8))&0xff;
}
// How to split the contents of val[8] which now holds the hex
return 0;
}
I am trying to split the hex value into 78661EB5, 4FE76763 and store each one inside an unsigned int which is stored inside val[8] in my example

Answer 1

Your variable long long hex is not a "char array", but you can try something like this, notice unsigned types throughout.
#include <stdio.h>
#include <stdint.h>
int main(void)
{
uint64_t hex = 0x78661EB54FE76763;
uint32_t lo, hi;
hi = hex >> 32;
lo = hex & 0xFFFFFFFF;
printf("%08X %08X\n", hi, lo);
return 0;
}
Program output:
78661EB5 4FE76763

Answer 2

Create a helper function that takes a pointer to an array of unsigned char and call it twice.
Left some things out for OP to discover.
unsigned long form_big_endian4(const unsigned char * x) {
unsigned long y = TBB; // What should the initial value be?
for (int i = 0; i < TBD; i++) { // How many times to loop?
y *= TBD; // Hint: max value of 8 bits + 1
y += x[i];
}
return y;
}
#include <stdio.h>
int main(void) {
const unsigned char val[8] =
{ 0x78, 0x66, 0x1E, 0xB5, 0x4F, 0xE7, 0x67, 0x63 };
unsigned long half = form_big_endian4(val); // Why long: Hint how small can unsigned be?
printf("%08lX\n", half); // Why 0 in format? Why `lX`?
half = form_big_endian4(val + TBD); // How far an offset?
printf("%08lX\n", half);
return 0;
}